An improved algorithm for integrated production and distribution scheduling problem with committed delivery dates

Abstract

In this paper, we study a machine scheduling problem with job transportation coordination. The orders including different amount of the same product will be delivered to their customers after being completed on a single machine and each order has a delivery due date which is promised by the manufacturer to its customer. A third-party logistics provider picks up the finished orders at the end of each day and charges each order a delivery cost, which is linearly decreasing with the delivery time and linearly increasing with its quantity. The goal is to find a schedule minimizing total delivery cost under the constraint that all the orders are delivered to the customers before the due dates. The problem is strongly NP-hard. We propose an improved approximate algorithm with worst-case ratio of $5/3$, which improves the existing one with worst-case ratio of $2$.

Appendices

Appendix A: Proof of Theorem 3.2

Lemma 5.1

If the heuristic runs step 2 and the obtained schedule is $\sigma _1$, then $\frac{f(\sigma _1)}{f^*}{\le }\frac{5}{3}$.

Proof

In schedule $\sigma _1$, it is clear to see that

$$\begin{aligned} f(\sigma _1)-f(\sigma _0)=aP_{1l}^{\sigma _1}~~~~and~~~~f^*\ge f(\sigma _0). \end{aligned}$$

(2)

(1) If $P_{1l}^{\sigma _{1}}=0$, then

$$\begin{aligned} \frac{f(\sigma _1)}{f^*}\le 1+\frac{f(\sigma _1)-f(\sigma _0)}{f(\sigma _0)}=1+\frac{aP_{1l}^{\sigma _1}}{f(\sigma _0)}=1. \end{aligned}$$

(2) If $d_{max}\ge d_{\sigma _{1}[1]}+1$ and $P_{1l}^{\sigma _{1}}\le \frac{2}{3}\sum _{i\in N}Q_i$, we have

$$\begin{aligned} b-a(d_{\sigma _{1}[1]}-1)>a~~~(by~~b-a(d_{max}-1)>0). \end{aligned}$$

This implies that

$$\begin{aligned} f(\sigma _0)&> ac+ \left( \sum _{i\in N}Q_i-c \right) [b-a(d_{max}-2)]\\&> ac+ \left( \sum _{i\in N}Q_i-c \right) a\\&= a\sum _{i\in N}Q_i. \end{aligned}$$

By $P_{1l}^{\sigma _{1}}\le \frac{2}{3}\sum _{i\in N}Q_i$, we have

$$\begin{aligned} \frac{f(\sigma _1)}{f^*}\le 1+\frac{f(\sigma _1)-f(\sigma _0)}{f(\sigma _0)}=1+\frac{aP_{1l}^{\sigma _1}}{f(\sigma _0)}\le 1+\frac{\frac{2a}{3}\sum _{i\in N}Q_i}{a\sum _{i\in N}Q_i}=\frac{5}{3}. \end{aligned}$$

(3) If $d_{max}=d_{\sigma _{1}[1]}$ and $P_{1l}^{\sigma _{1}}\le \frac{2}{3} \left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\Delta x \right)$, by $b-a(d_{max}-1)>0$, the unit shipping cost for products delivered at the end of day $2$ is at least $a$. Thus, $f(\sigma _0)>a \left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\Delta x \right)$ and

$$\begin{aligned} \frac{f(\sigma _1)}{f^*}&\le 1+\frac{f(\sigma _1)-f(\sigma _0)}{f(\sigma _0)}\\&= 1+\frac{aP_{1l}^{\sigma _1}}{f(\sigma _0)}\\&< 1+\frac{\frac{2}{3}a \left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\Delta x \right) }{a \left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\Delta x \right) }\\&= \frac{5}{3}. \end{aligned}$$

We can conclude that, if the output schedule is $\sigma _1$, then $\frac{f(\sigma _1)}{f^*}\le \frac{5}{3}$. $\square$

Lemma 5.2

If the heuristic runs step 2 and the output schedule is $\sigma _2^{P2}$, then $\frac{f(\sigma _2^{P2})}{f^*}\le \frac{5}{3}$.

Proof

By $P_{1l}^{\sigma _{1}}>\frac{2}{3}\sum _{i\in N}Q_i$, we know that $\sigma _{1}[1]$ is the largest order in $N$. Thus, the structure of $\sigma _1$(and $\sigma _0$) is ($N_1, N_2, \sigma _{1}[1], N_4$), and the structure of $\sigma _2^{P2}$ is ($N_1, \sigma _{1}[1], N_4, N_2$). Therefore,

$$\begin{aligned} f(\sigma _2^{P2})-f^*\le a\sum _{i\in N_2}Q_i+a\sum _{i\in N_4}Q_i<\frac{a}{3}\sum _{i\in N}Q_i. \end{aligned}$$

As discussed in the proof of Lemma 5.1, $f(\sigma _0)>a\sum _{i\in N}Q_i$ and

$$\begin{aligned} \frac{f(\sigma _2^{P2})}{f^*}\le 1+\frac{f(\sigma _2^{P2})-f^*}{f(\sigma _0)}<1+\frac{\frac{a}{3}\sum _{i\in N}Q_i}{f(\sigma _0)}<1+\frac{\frac{a}{3}\sum _{i\in N}Q_i}{a\sum _{i\in N}Q_i}<\frac{5}{3}. \end{aligned}$$

$\square$

Lemma 5.3

If the heuristic runs step 2 and the output schedule is $\sigma _3^{P2}$, then $\frac{f(\sigma _3^{P2})}{f^*}\le \frac{5}{3}$.

Proof

In this case, the following three conditions hold:

• (1) $d_{max}=d_{\sigma _{1}[1]}$;

• (2) $P_{1l}^{\sigma _{1}}>\frac{2}{3} \left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\Delta x \right)$;

• (3) $P_{1r}^{\sigma _{1}}>\sum _{i\in N_2}Q_i$.

Note that in schedule $\sigma _1$, $\Delta$ is the set of orders processed between $N_2$ and $\sigma _1[1]$ and $Q_{max}$ = max$\{Q_j|j\in \Delta \}$, then the structure of $\sigma _1$(and $\sigma _0$) is ($N_1, N_2, \Delta , \sigma _{1}[1], N_4$) and the structure of $\sigma _3^{P2}$ is ($N_1, \Delta , N_4, N_2, \sigma _{1}[1]$).

We first consider case (1): $\sum _{i\in N_4}Q_i\le P_{1l}^{\sigma _{1}}$. We prove that, in this case, $\sigma _{3}^{P2}$ is an optimal schedule. By $\sum _{i\in N_4}Q_i\le P_{1l}^{\sigma _{1}}$, we can see that order $\sigma _{1}[1]$ is the only order that is finished production on the second day in schedule $\sigma _3^{P2}$. Thus, along with the fact that $d_j=d_{\sigma _1[1]}$ and $Q_j\ge Q_{\sigma _1[1]}$ for $j\in \Delta$ (By EDD-LPT), we can conclude that any feasible schedule where an order from $\Delta$ is finished on day $2$ will have a total delivery cost greater than or equal to $f(\sigma _3^{P2})$. Now consider any feasible schedule $\pi$ where all the orders in $\Delta$ are finished on day $1$. The orders in $N_1$ must be finished on day $1$ in $\pi$. Due to the fact that $P_{1r}^{\sigma _{1}}>\sum _{i\in N_2}Q_i$, even if all the other orders, i.e. orders in $N_2\bigcup N_4$, are finished production on day 2, order $\sigma _1[1]$ will still be finished on day $2$. Thus, the total cost of $\pi$ is at least $f(\sigma _3^{P2})$. Based on the above analysis, we can see that $\sigma _3^{P2}$ is an optimal schedule.

Next we consider case (2): $\sum _{i\in N_4}Q_i>P_{1l}^{\sigma _{1}}$ and $S_4\ne \emptyset$. By the definitions of $S_4$ and $N_4^*$, we can see that the structure of $\sigma _3^{P2}$ is ($N_1, \Delta , S_4, \sigma _3^{P2}[1], N_4^*, N_2, \sigma _1[1]$) and $f(\sigma _3^{P2})-f(\sigma _1)\le a\sum _{i\in N_2}Q_i-a\sum _{i\in S_4}Q_i$. Note that $S_4\ne \emptyset$ and $Q_j\ge Q_{\sigma _3^{P2}[1]}$ for $j\in S_4$ (By EDD-LPT), then $Q_{\sigma _3^{P2}[1]}<\frac{1}{2}\sum _{i\in N_4}Q_i$ and $\sum _{i\in S_4}Q_i>\frac{1}{2}P_{1l}^{\sigma _1}$. Thus,

$$\begin{aligned} f(\sigma _3^{P2})-f(\sigma _0)&= f(\sigma _3^{P2})-f(\sigma _1)+f(\sigma _1)-f(\sigma _0)\\&\le a\sum _{i\in N_2}Q_i-a\sum _{i\in S_4}Q_i+aP_{1l}^{\sigma _1}\\&< a\sum _{i\in N_2}Q_i+\frac{a}{2}P_{1l}^{\sigma _1}. \end{aligned}$$

By condition $\sum _{i\in N_4}Q_i>P_{1l}^{\sigma _{1}}$, we have

$$\begin{aligned} \Delta x-P_{1l}^{\sigma _{1}}>\Delta x-\sum _{i\in N_4}Q_i=P_{1r}^{\sigma _1}>\sum _{i\in N_2}Q_i \end{aligned}$$

and $\Delta x>P_{1l}^{\sigma _{1}}+\sum _{i\in N_2}Q_i$. By condition (1), it is clear to see that

$$\begin{aligned} f(\sigma _0)>a \left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\Delta x \right) \end{aligned}$$

and

$$\begin{aligned} \frac{f(\sigma _3^{P2})}{f^*}&\le 1+\frac{f(\sigma _3^{(P2)})-f(\sigma _0)}{f(\sigma _0)}\\&< 1+\frac{a\sum _{i\in N_2}Q_i+\frac{a}{2}P_{1l}^{\sigma _1}}{a(\sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\Delta x)}\\&< 1+\frac{a\sum _{i\in N_2}Q_i+\frac{a}{2}P_{1l}^{\sigma _1}}{a(P_{1l}^{\sigma _{1}}+2\sum _{i\in N_2}Q_i)}\\&< \frac{5}{3}. \end{aligned}$$

At last we consider case (3): $\sum _{i\in N_4}Q_i>P_{1l}^{\sigma _{1}},~~S_4=\emptyset$ and $Q_{max}+\sum _{i\in N_2}Q_i+\sum _{i\in N_4^*}Q_i<\Delta x.$ In $\sigma _3^{P2}$, we can see that the orders in $N_4^*$ and $N_2$ are produced on day 2, then at least two orders in $N_{max}$ are finished on day $2$ in any feasible schedule. Along with the fact that $d_j=d_{\sigma _1[1]}=d_{\sigma _3^{P2}[1]}$, $Q_j\ge Q_{\sigma _1[1]}\ge Q_{\sigma _3^{P2}[1]}$ for $j\in \Delta$ (By EDD-LPT) and the definition of $N_{max}$, we have

$$\begin{aligned} f(\sigma _3^{P2})-f^*<a\sum _{i\in N_2}+a\sum _{i\in N_4^*}Q_i. \end{aligned}$$

Recall that $N_4=S_4\bigcup \sigma _3^{P2}[1]\bigcup N_4^*$ and $|N_4|<\Delta x$, then

$$\begin{aligned} Q_{\sigma _3^{P2}[1]}&> P_{1l}^{\sigma _3^{P2}}\\&\ge P_{1l}^{\sigma _1}\\&> \frac{2}{3}\left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\Delta x \right) \\&> \frac{2}{3}\left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+Q_{\sigma _3^{P2}[1]}+\sum _{i\in N_4^*}Q_i \right) , \end{aligned}$$

which indicates that $Q_{\sigma _3^{P2}[1]}>2 \left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\sum _{i\in N_4^*}Q_i \right)$. Thus, we have

$$\begin{aligned} \frac{f(\sigma _3^{P2})}{f^*}&\le 1+\frac{f(\sigma _3^{P2})-f(\sigma _0)}{f(\sigma _0)}\\&< 1+\frac{a\sum _{i\in N_2}Q_i+a\sum _{i\in N_4^*}Q_i}{a \left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\Delta x \right) }\\&\le 1+\frac{a\sum _{i\in N_2}Q_i+a\sum _{i\in N_4^*}Q_i}{a \left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\sum _{i\in N_4^*}Q_i+Q_{\sigma _3[1]} \right) }\\&< 1+\frac{a\sum _{i\in N_2}Q_i+a\sum _{i\in N_4^*}Q_i}{3a \left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\sum _{i\in N_4^*}Q_i \right) }\\&< \frac{5}{3}. \end{aligned}$$

Therefore, if the output schedule is $\sigma _3^{P2}$, then $\frac{f(\sigma _3^{P2})}{f^*}\le \frac{5}{3}$. $\square$

Lemma 5.4

If the heuristic runs step 2 and the output schedule is $\sigma _4^{P2}$, then $\frac{f(\sigma _4^{P2})}{f^*}\le \frac{5}{3}$.

Proof

Note that $\sum _{i\in N_2}Q_i+\sum _{i\in N_4^*}Q_i<P_{1r}^{\sigma _1}+N_4^*<\Delta x$ and $P_{1l}^{\sigma _1}>\frac{2}{3}\Delta x$, we can see that in any feasible schedule, there is one or two orders in $N_{max}$ being finished on day 2.

Now consider two possible cases as follows:

(i) If only one order in $N_{max}$ is finished on day $2$ in the optimal schedule, then

$$\begin{aligned} f(\pi _2)-f^*<a\sum _{i\in N_2}+a\sum _{i\in N_4^*}Q_i. \end{aligned}$$

As discussed in the proof of Lemma 5.3, we have $\frac{f(\pi _2)}{f^*}\le \frac{5}{3}$.

(ii) If two orders in $N_{max}$ are finished on day $2$ in the optimal schedule, since $d_i=d_j=d_{\sigma _1[1]}=d_{\sigma _3[1]}$ and

$$\begin{aligned} Q_i\ge Q_{\sigma _1[1]}\ge Q_{\sigma _3^{P2}[1]},~~Q_j\ge Q_{\sigma _1[1]}\ge Q_{\sigma _3^{P2}[1]}~~for~~i,~j\in N_{max}\backslash \{Q_{\sigma _3^{P2}[1]}\}, \end{aligned}$$

we can see that $Q_i+Q_j\ge Q_{\sigma _1[1]}+Q_{\sigma _3[1]}$. Thus, any feasible schedule where two orders from $N_{max}$ are finished on day $2$ will have a total cost greater than or equal to $f(\pi _1)$. Based on the above analysis, we can see that $\pi _1$ is an optimal schedule in this case.

Therefore, if the output schedule is $\sigma _4^{P2}$, then $\frac{f(\sigma _4^{P2})}{f^*}\le \frac{5}{3}$. $\square$

Lemma 5.5

If the heuristic runs step 2 and the output schedule is $\sigma _5^{P2}$, then $\frac{f(\sigma _5^{P2})}{f^*}\le \frac{5}{3}$.

Proof

We define:

• $S_2$ $=$ the set of orders which are completed on day $1$ in $N_2$ in schedule $\sigma _5^{P2}$.

• $N_2^*$ $=$ the set of orders which are completed on day $2$ in $N_2$ in schedule $\sigma _5^{P2}$.

As discussed in the proof of Lemma 5.4, we can see that the structure of $\sigma _{1}$(and $\sigma _{0}$) is $(N_1, N_2, \Delta , \sigma _1[1], N_4)$. By the definitions of $S_2$ and $N_2^*$, the structure of $\sigma _5^{P2}$ is $(N_1, \Delta , \sigma _1[1], S_2, N_2^*, N_4)$. Now consider two possible cases as follows:

Case (1) If $S_2\ne \emptyset$, then we have $f(\sigma _5^{P2})-f(\sigma _1)=-aQ_{\sigma _1[1]}+a\sum _{i\in N_2^*}Q_i$ and

$$\begin{aligned} f(\sigma _5^{P2})-f(\sigma _0)&= f(\sigma _5^{P2})-f(\sigma _1)+f(\sigma _1)-f(\sigma _0)\\&= a\sum _{i\in N_2^*}Q_i-a Q_{\sigma _1[1]}+a P_{1l}^{\sigma _1}\\&= a\sum _{i\in N_2^*}Q_i-a P_{1r}^{\sigma _1}\\&= a P_{1l}^{\sigma _5^{P2}}. \end{aligned}$$

Since $S_2\ne \emptyset$ and $Q_j\ge Q_{\sigma _5^{P2}[1]}$ for $j\in S_2$ (By LPT), $P_{1l}^{\sigma _5^{P2} }<Q_{\sigma _5^{P2}[1]}<\frac{1}{2}\sum _{i\in N_2}Q_i$ and

$$\begin{aligned} \frac{f(\sigma _5^{P2})}{f^*}&\le 1+\frac{f(\sigma _5^{P2})-f(\sigma _0)}{f(\sigma _0)}\\&< 1+\frac{aP_{1l}^{\sigma _5^{P2}}}{a \left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\Delta x \right) }\\&< 1+\frac{\frac{a}{2}\sum _{i\in N_2}Q_i}{a\sum _{i\in N_2}Q_i}\\&< \frac{5}{3}. \end{aligned}$$

Case (2) If $S_2=\emptyset$, then we consider the following two subcases:

(i) If order $\sigma _5^{P2}[1]$ is finished production on day 1 in the optimal schedule, then at least one order in $\{\sigma _1[1]\}\bigcup \Delta$ must be finished on day 2 and $f^*>aQ_{\sigma _1[1]}+aP_{1l}^{\sigma _5^{P2}}>\frac{5}{3}aP_{1l}^{\sigma _5^{P2}}$(by $P_{1l}^{\sigma _{1}}>\frac{2}{3}\left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\Delta x\right)$. Thus,

$$\begin{aligned} \frac{f(\sigma _5^{P2})}{f^*}\le 1+\frac{f(\sigma _5^{P2})-f(\sigma _0)}{f^*}<1+\frac{aP_{1l}^{\sigma _5^{P2}}}{\frac{5}{3}aP_{1l}^{\sigma _5^{P2}}}<\frac{5}{3}. \end{aligned}$$

(ii) If order $\sigma _5^{P2}[1]$ is finished production on day 2 in the optimal schedule, then we have $f^*>2aP_{1l}^{\sigma _5^{P2}}$ and

$$\begin{aligned} \frac{f(\sigma _5^{P2})}{f^*}\le 1+\frac{f(\sigma _5^{P2})-f(\sigma _0)}{f^*}\le 1+\frac{aP_{1l}^{\sigma _5^{P2}}}{2aP_{1l}^{\sigma _5^{P2}}}<\frac{5}{3}. \end{aligned}$$

Therefore, $\frac{f(\sigma _5^{P2})}{f^*}<\frac{5}{3}$. $\square$

Theorem 3.2 If the output schedule of heuristic $\mathbf {H}$ is obtained from step 2, then $\frac{f(\sigma ^{\mathbf {H}})}{f^*}\le 5/3$.

Proof

This is a direct conclusion from Lemmas 5.1 through 5.5. $\square$

Appendix B: Proof of Theorem 3.3

Lemma 6.1

If the heuristic runs step 3 and the obtained schedule is $\sigma _1$, then $\frac{f(\sigma _1)}{f^*}\le \frac{5}{3}$.

Proof

In schedule $\sigma _1$, it is clear to see that $f(\sigma _1)-f(\sigma _0)=a(P_{1l}^{\sigma _1}+P_{2l}^{\sigma _1})<2ac$.

(1) If $d_{\sigma _{1}[2]}\le d_{max}-1$, then by $b-a(d_{max}-1)>0$, we have

$$\begin{aligned} f^*\ge f(\sigma _0)&> c[b-a(d_{\sigma _{1}[2]}-1)]+c[b-a(d_{\sigma _{1}[2]}-2)]\\&> c[b-a(d_{max}-1-1)]+c[b-a(d_{max}-1-2)]\\&> ac+2ac=3ac \end{aligned}$$

and

$$\begin{aligned} \frac{f(\sigma _1)}{f^*}\le 1+\frac{f(\sigma _1)-f(\sigma _0)}{f(\sigma _0)}<1+\frac{2ac}{3ac}=\frac{5}{3}. \end{aligned}$$

(2) If $d_{\sigma _{1}[1]}\le d_{max}-2$, then by the same analysis as that in the proof of (1), we have

$$\begin{aligned} \frac{f(\sigma _1)}{f^*}\le \frac{5}{3}. \end{aligned}$$

(3) If $d_{max}=d_{\sigma _{1}[1]}+1$ and $P_{1l}^{\sigma _{1}}+P_{2l}^{\sigma _{1}}\le \frac{2}{3}(2c+2\Delta x)$, then

$$\begin{aligned} f(\sigma _0)&> c[b-a(d_{\sigma _{1}[1]}-1)]+c[b-a(d_{\sigma _{1}[2]}-2)]+\Delta x[b-a(d_{max}-3)]\\&> c[b-a(d_{max}-1-1)]+c[b-a(d_{max}-2)]+\Delta x[b-a(d_{max}-3)]\\&> 2ac+2a\Delta x. \end{aligned}$$

By $P_{1l}^{\sigma _{1}}+P_{2l}^{\sigma _{1}}\le \frac{2}{3}(2c+2\Delta x)$, we have

$$\begin{aligned} \frac{f(\sigma _1)}{f^*}&\le 1+\frac{f(\sigma _1)-f(\sigma _0)}{f(\sigma _0)}\\&= 1+\frac{a(P_{1l}^{\sigma _1}+P_{2l}^{\sigma _1})}{f(\sigma _0)}\\&\le 1+\frac{\frac{2a}{3}(2c+2\Delta x)}{2ac+2a\Delta x}\\&= \frac{5}{3}. \end{aligned}$$

(4) If $d_{max}=d_{\sigma _{1}[1]}+1=4$ and $\Delta _2=\emptyset$, then the structure of $\sigma _1$(and $\sigma _0$) is ($N_1,~\Delta _1,~\sigma _{1}[1],~N_5,~\sigma _{1}[2],~N_4$). Let $Q_{\Delta _1}=$ $\sum _{j\in _{\Delta _1}}Q_j$, then $\sum _{i\in N_1\bigcup N_5}Q_i+Q_{\Delta _1}+Q_{\sigma _{1}[1]}+Q_{\sigma _{1}[2]}>2c.$ By the definitions of $N_1$, $N_5$ and $\Delta _1$ and $d_{max}=d_{\sigma _{1}[1]}+1=4$, we can see that the orders in $N_1$ must be finished production by the end of day 1 and the orders in $N_5\bigcup \{\sigma _1[1]\}\bigcup \Delta _1$ must be finished production by the end of day 2. Therefore, $\sigma _1[2]$ must be finished production on day 3. By Remark 3.1 , we can see that if $\Delta _1\ne \emptyset$, then $\Delta _1$ contains only one order and $Q_{\Delta _1}\ge Q_{\sigma _{1}[1]}$ (by EDD-LPT). Since $\sum _{i\in N_1\bigcup \Delta _1}Q_i<c$,

$$\begin{aligned} f(\sigma _{1})-f^*\le a\sum _{i\in N_5}Q_i+2a\sum _{i\in N_4}Q_i \end{aligned}$$

and

$$\begin{aligned} \sum _{i\in N_4\bigcup N_5}&< \sum _{i\in N}Q_i-P_{1l}^{\sigma _{1}}-P_{2l}^{\sigma _{1}}\\&< 2c+\Delta x-\frac{2}{3}(2c+2\Delta x)\\&< \frac{2c}{3}. \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{f(\sigma _1)}{f^*}&= 1+\frac{f(\sigma _1)-f^*}{f^*}\\&< 1+\frac{a\sum _{i\in N_4}Q_i+2a\sum _{i\in N_5}Q_i}{2ac}\\&< 1+\frac{\frac{4ac}{3}}{2ac}=\frac{5}{3}. \end{aligned}$$

(5) If $d_{max}=d_{\sigma _{1}[1]}$ and $P_{1l}^{\sigma _{1}}+P_{2l}^{\sigma _{1}}\le \frac{2}{3} \big (2\sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+c+2\Delta x \big )$, by the same analysis as that in the proof of (3), we have $f(\sigma _0)\!>\!a \left( 2\sum _{i\in N_1}Q_i\!+\!\sum _{i\in N_2}Q_i\!+\!\sum _{i\in N_3}Q_i\!+\!c\!+\!2\Delta x \right)$ and $\frac{f(\sigma _1)}{f^*}\le \frac{5}{3}$.

We can conclude that, if the output schedule is $\sigma _1$, then $\frac{f(\sigma _1)}{f^*}\le \frac{5}{3}$. $\square$

Lemma 6.2

If the heuristic runs step 3 and the obtained schedule is $\sigma _2^{P3}$, then $\frac{f(\sigma _2^{P3})}{f^*}\le \frac{5}{3}$.

Proof

In this case, the following three conditions hold:

• (1) $d_{max}=d_{\sigma _{1}[1]}+1=d_{\sigma _{1}[2]}=4$;

• (2) $P_{1l}^{\sigma _{1}}+P_{2l}^{\sigma _{1}}>\frac{2}{3}(2c+2\Delta x)$;

• (3) $\Delta _2\ne \emptyset$.

By Remark 3.1 and condition (3), we can see that $\Delta _1=\emptyset$ and $\Delta _2$ contains only one order. The structure of $\sigma _1$ (and $\sigma _0$) is ($N_1,~\sigma _{1}[1],~N_5,~\Delta _2,~\sigma _{1}[2],~N_4$) and the structure of $\sigma _2^{P3}$ is ($N_1,~\Delta _2,~\sigma _{1}[1],~N_5,~\sigma _{1}[2],~N_4$). Let $Q_{\Delta _2}=$ $\sum _{j\in _{\Delta _2}}Q_j$. By conditions (2) and (3), we have $\frac{c}{3}<P_{2l}^{\sigma _1}<\frac{c}{2}$, $P_{1l}^{\sigma _1}>\frac{5c}{6}>\frac{2c}{3}>Q_{\Delta _2}$ and $\sum _{i\in N_1\bigcup \Delta _2}Q_i<c$. Since $\Delta _1=\emptyset$, $\sum _{i\in N_1}Q_i+Q_{\sigma _{1}[1]}>c$. By the same analysis as that of case (4) in Lemma 6.1 , we know that $\sigma _1[2]$ must be finished production on day 2 and

$$\begin{aligned} f(\sigma _{1})-f^*\le a\sum _{i\in N_5}Q_i+2a\sum _{i\in N_4}Q_i. \end{aligned}$$

By the same analysis as that of case (4) in Lemma 6.1, we have $\frac{f(\sigma _2^{P3})}{f^*}\le \frac{5}{3}$. $\square$

Lemma 6.3

If the heuristic runs step 3 and the obtained schedule is $\sigma _3^{P3}$, then $\frac{f(\sigma _3^{P3})}{f^*}\le \frac{5}{3}$.

Proof

It is easy to check that schedules $\pi _{1}\sim \pi _{12}$ defined in Step 3 of the algorithm are all feasible. By the definitions of $N_1$ and $N_2$, the orders in $N_1$ must be finished production by the end of day 1 and the orders in $N_2$ must be finished production by the end of day 2. By Remark 3.1, we can see that if $\Delta _1\ne \emptyset$ , then $\Delta _2=\emptyset$ and $\Delta _1$ contains only one order. If $\Delta _2\ne \emptyset$, then $\Delta _1=\emptyset$ and $\Delta _2$ contains only one order. Therefore,

$$\begin{aligned} f(\sigma _{3}^{P3})-f^*\le a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3\bigcup N_5\bigcup N_4}Q_i. \end{aligned}$$

By the same analysis as that of case (4) in Lemma 6.1, we have $\frac{f(\sigma _2^{P3})}{f^*}\le \frac{5}{3}$. $\square$

Lemma 6.4

If the heuristic runs step 3 and the output schedule is $\sigma _4^{P3}$, then $\frac{f(\sigma _4^{P3})}{f^*}\le \frac{5}{3}$.

Proof

In this case, the following three conditions hold:

• (1) $d_{max}=d_{\sigma _{1}[1]}=d_{\sigma _{1}[2]}\ge 4$;

• (2) $P_{1l}^{\sigma _{1}}+P_{2l}^{\sigma _{1}}>\frac{2}{3} \left( 2\sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+c+2\Delta x \right)$;

• (3) $N_3\ge$max$\{P_{1r}^{\sigma _1},~P_{2r}^{\sigma _1}\}$.

By condition (3), we know that $\sigma _1[1]$ is finished production on day 1 and $\sigma _1[2]$ is finished production on day 2 in schedule $\sigma ^{P3}_4$, $f(\sigma _4^{P3})-f(\sigma _1)<-a(Q_{\sigma _1[1]}+Q_{\sigma _1[2]})+2a\sum _{i\in N_3}Q_i$ and $f(\sigma _4^{P3})-f(\sigma _0)<2a\sum _{i\in N_3}Q_i-P_{2r}^{\sigma _1}-P_{1r}^{\sigma _1}.$

Let $S_3$ = the set of orders in $N_3$ which are completed on day $2$ in schedule $\sigma _4^{P3}$ and $Q_{3,max}=$max$\{Q_j|j\in N_3\}$. Now we consider three possible cases:

Case 4.1: If $S_3\ne \emptyset$, we have

$$\begin{aligned} f(\sigma _4^{P3})-f(\sigma _1)&< -a(Q_{\sigma _1[1]}+Q_{\sigma _1[2]})+\frac{a}{2} \left( \sum _{i\in N_3}Q_i-P_{2r}^{\sigma _1} \right) \\&+\,2a \left[ \sum _{i\in N_3}Q_i-\frac{1}{2} \left( \sum _{i\in N_3}Q_i-P_{2r}^{\sigma _1}\right) \right] \\&< -a(Q_{\sigma _1[1]}+Q_{\sigma _1[2]})+\frac{3a}{2}\sum _{i\in N_3}Q_i+\frac{a}{2}P_{2r}^{\sigma _1}\\ \end{aligned}$$

and

$$\begin{aligned} f(\sigma _4^{P3})-f(\sigma _0)<-aP_{2r}^{\sigma _1}+\frac{3a}{2}\sum _{i\in N_3}Q_i+\frac{a}{2}P_{2r}^{\sigma _1}<\frac{3a}{2}\sum _{i\in N_3}Q_i. \end{aligned}$$

By condition (2) and $P_{2l}^{\sigma _1}<c$, we have $P_{1l}^{\sigma _1}>\frac{2}{3}\sum _{i\in N_3}Q_i-\frac{c}{3}$. Note that $P_{1l}^{\sigma _1}+\sum _{i\in N_3}Q_i\le c$, thus $\frac{2}{3}\sum _{i\in N_3}Q_i-\frac{c}{3}+\sum _{i\in N_3}Q_i\le c$ and $c>\frac{5}{4}\sum _{i\in N_3}Q_i$. Since $f^*>a \left( c+\sum _{i\in N_3}Q_i \right) >\frac{9}{4}a\sum _{i\in N_3}Q_i$, $\frac{f(\sigma _4^{P3})}{f^*}\le \frac{5}{3}$.

Case 4.2: If $S_3=\emptyset$ and $\sum _{i\in N_3}Q_i\le Q_{\sigma _1[2]}$, we consider the following subcases:

4.2.1 $Q_{3,max}$ is finished production on day 1 in the optimal schedule. By $S_3=\emptyset$, we can see that at least one order $i$ with $Q_i\ge Q_{\sigma _{1}[2]}$ is finished production on day 3 in this optimal solution. Thus, $f^*>aQ_{3,max}+a \left( \sum _{i\in N_3}Q_i-Q_{3,max} \right) +2aQ_{\sigma _1[2]}>3a\sum _{i\in N_3}Q_i$. Note that $f(\sigma _4^{P3})-f(\sigma _0)\le 2a\sum _{i\in N_3}Q_i$, then $\frac{f(\sigma _4^{P3})}{f^*}\le \frac{5}{3}$.

4.2.2 $Q_{3,max}$ is finished production on day 2 in the optimal schedule. By $S_3=\emptyset$, we can see that at least one order $i$ with $Q_i\ge Q_{\sigma _{1}[2]}$ is finished production on day 3 in this optimal schedule. Thus, $f^*>2aQ_{3,max}+a \left( \sum _{i\in N_3}Q_i-Q_{3,max} \right) +2aQ_{\sigma _1[2]}>3a\sum _{i\in N_3}Q_i$ and $\frac{f(\sigma _4^{P3})}{f^*}\le \frac{5}{3}$.

4.2.3 $Q_{3,max}$ is finished production on day 3 in the optimal schedule. We can see that at least one order $i$ with $Q_i\ge Q_{\sigma _{1}[2]}$ is finished production on day 2 in the optimal solution. Therefore, $f^*>3aQ_{3,max}+a \left( \sum _{i\in N_3}Q_i-Q_{3,max}\right) +aQ_{\sigma _1[2]}>2a\sum _{i\in N_3}Q_i+2aQ_{3,max}$. Note that $f(\sigma _4^{P3})-f(\sigma _0)\le 2a\sum _{i\in N_3}Q_i-(P_{1r}^{\sigma _1}+P_{2r}^{\sigma _1})<a\sum _{i\in N_3}Q_i+Q_{3,max}$, then $\frac{f(\sigma _4^{P3})}{f^*}\le \frac{5}{3}$.

Case 4.3: If $S_3=\emptyset$ and $\sum _{i\in N_3}Q_i>Q_{\sigma _1[2]}$, then by condition (2) and $P_{1l}^{\sigma _1}\le c-\sum _{i\in N_3}Q_i$, we have

$$\begin{aligned} \frac{2}{3} \left( \sum _{i\in N_3}Q_i+c \right)&< \frac{2}{3} \left( 2\sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+c+2\Delta x \right) \\&< P_{1l}^{\sigma _{1}}+P_{2l}^{\sigma _{1}}\\&< c-\sum _{i\in N_3}Q_i+\sum _{i\in N_3}Q_i. \end{aligned}$$

Therefore, $c>2\sum _{i\in N_3}Q_i$ and $f^*>c+\sum _{i\in N_3}Q_i>3\sum _{i\in N_3}Q_i$. Note that $f(\sigma _4^{P3})-f(\sigma _0)\le 2a\sum _{i\in N_3}Q_i$, then $\frac{f(\sigma _4^{P3})}{f^*}\le \frac{5}{3}$. $\square$

Lemma 6.5

If the heuristic runs step 3 and the output schedule is $\sigma _5^{P3}$, then $\frac{f(\sigma _5^{P3})}{f^*}\le \frac{5}{3}$.

Proof

In this case, the following three conditions hold:

• (1) $d_{max}=d_{\sigma _{1}[1]}=d_{\sigma _{1}[2]}\ge 4$;

• (2) $P_{1l}^{\sigma _{1}}+P_{2l}^{\sigma _{1}}>\frac{2}{3} \left( 2\sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+c+2\Delta x \right)$;

• (3) $N_3<$max$\{P_{1r}^{\sigma _1},~~P_{2r}^{\sigma _1}\}$.

Now we consider the following possible cases:

Case 5.1: If $\sigma _1[1]$ is completed on day 1 and $\sigma _1[2]$ is completed on day 2 in schedule $\pi _{13}$, then

$$\begin{aligned} f(\pi _{13})-f(\sigma _1)<-a(Q_{\sigma _1[1]}+Q_{\sigma _1[2]})+a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i \end{aligned}$$

and

$$\begin{aligned} f(\pi _{13})-f(\sigma _0)<-a(P_{1r}^{\sigma _1}+P_{2r}^{\sigma _1})+a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i. \end{aligned}$$

By condition (3), we have $f(\pi _{13})-f(\sigma _0)<a\sum _{i\in N_2}Q_i+a\sum _{i\in N_3}Q_i.$ Note that

$$\begin{aligned} f^*>ac+a\sum _{i\in N_2}Q_i+a\sum _{i\in N_3}Q_i>2\left( a\sum _{i\in N_2}Q_i+a\sum _{i\in N_3}Q_i\right) , \end{aligned}$$

thus, $\frac{f(\pi _{13})}{f^*}\le \frac{5}{3}$.

Case 5.2: If $\sigma _1[1]$ is completed on day 1 and $\sigma _1[2]$ is completed on day 3 in schedule $\pi _{13}$, then $\sum _{i\in N_3}<P_{2r}^{\sigma _1}$ and

$$\begin{aligned} \sum _{i\in N_1\bigcup \Delta _1\bigcup \Delta _2\bigcup N_2}Q_i+Q_{\sigma _1[1]}+Q_{\sigma _1[2]}>2c. \end{aligned}$$

By EDD-LPT, we have $d_j=d_{max}=d_{\sigma _1[2]}$ and $Q_j\ge Q_{\sigma _1[2]}$ for $j\in \{\sigma _1[1]\}\bigcup \Delta _1\bigcup \Delta _2$. By the definitions of $N_1$ and $N_2$, the orders in $N_1$ must be finished production by the end of day 1 and the orders in $N_2$ must be finished production by the end of day 2. Consequently, there must be one order $i$ with $Q_i\ge Q_{\sigma _1[2]}$ being completed on day 3. Now we consider two subcases:

Case 5.2.1: If $d_{max}=4$, then $N_3=\emptyset$. By the same analysis as that of case 5.1, we have $f(\pi _{13})-f(\sigma _0)<aP_{2l}^{\sigma _1}+a\sum _{i\in N_2}Q_i$. Note that $f^*>a\sum _{i\in N_2}Q_i+aQ_{\sigma _1[2]}+c>\frac{3}{2} \left( a\sum _{i\in N_2}Q_i+aQ_{\sigma _1[2]}\right)$, then $\frac{f(\pi _{13})}{f^*}\le \frac{5}{3}$.

Case 5.2.2: If $d_{max}\ge 5$, then by $\sum _{i\in N_3}<P_{2r}^{\sigma _1}$,

$$\begin{aligned} f^*&> 2a\sum _{i\in N_2}Q_i+2aQ_{\sigma _1[2]}+a\sum _{i\in N_3}Q_i\\&= 2a\sum _{i\in N_2}Q_i+2aP_{2l}^{\sigma _1}+2aP_{2r}^{\sigma _1}+a\sum _{i\in N_3}Q_i\\&> 2a\sum _{i\in N_2}Q_i+2aP_{2l}^{\sigma _1}+3a\sum _{i\in N_3}Q_i\\&> \frac{3}{2}a \left( P_{2l}^{\sigma _1}+\sum _{i\in N_2}Q_i+2\sum _{i\in N_3}Q_i \right) . \end{aligned}$$

By the same analysis as that of case 5.1, $f(\pi _{13})-f(\sigma _0)<aP_{2l}^{\sigma _1}+a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i$ and $\frac{f(\pi _{13})}{f^*}\le \frac{5}{3}$.

Case 5.3: If $\sigma _1[1]$ is completed on day 2 in schedule $\pi _{13}$, then (1) $\sum _{i\in N_1\bigcup \Delta _1}Q_i+Q_{\sigma _1[1]}>c$, (2) there must be one order $i\in \{\sigma _1[1]\}\bigcup \Delta _1$ with $Q_i\ge Q_{\sigma _1[1]}$ being completed on day 2, and (3) $\sum _{i\in N_2\bigcup N_3}Q_i<P_{1r}^{\sigma _1}$.

Let $Q_{\Delta _2,max}=$max$\{Q_j|j\in \Delta _2\}$ and $S_{\Delta _2}$ = the set of orders from $\Delta _2$ which are completed on day $1$ in schedule $\pi _{14}$. Now we consider the following possible subcases:

Case 5.3.1: If $S_{\Delta _2}\ne \emptyset$ and $\sigma _1[2]$ is completed on day 2 in schedule $\pi _{14}$, then

$$\begin{aligned} f(\pi _{14})-f(\sigma _1)<-\frac{a}{2} \left( \sum _{i\in N_2\bigcup N_3}Q_i+P_{1l}^{\sigma _1} \right) +a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i-aQ_{\sigma _1[2]} \end{aligned}$$

and

$$\begin{aligned} f(\pi _{14})-f(\sigma _0)&< -\frac{a}{2}\left( \sum _{i\in N_2\bigcup N_3}Q_i+P_{1l}^{\sigma _1}\right) +a\sum _{i\in N_2}Q_i\\&+2a\sum _{i\in N_3}Q_i-a Q_{\sigma _1[2]}+a(P_{1l}^{\sigma _1}+P_{2l}^{\sigma _1})\\&< \frac{a}{2} \left( P_{1l}^{\sigma _1}+\sum _{i\in N_2}Q_i \right) +\frac{3a}{2}\sum _{i\in N_3}Q_i. \end{aligned}$$

Note that $S_{\Delta _2}\ne \emptyset$, then $\Delta _2\ne \emptyset$ and $P_{2l}^{\sigma _1}<\frac{c}{2}$. By $c>\sum _{i\in N_3}Q_i+P_{1l}^{\sigma _1}$ and condition (2), we have

$$\begin{aligned} c-\sum _{i\in N_3}Q_i+\frac{c}{2}&> P_{1l}^{\sigma _1}+P_{2l}^{\sigma _1}\\&> \frac{2}{3}\left( \sum _{i\in N_3}Q_i+c \right) \\ \end{aligned}$$

and $c>2\sum _{i\in N_3}Q_i$. Therefore,

$$\begin{aligned} f^*&> a\sum _{i\in N_2\bigcup N_3}Q_i+ac\\&= a\sum _{i\in N_2\bigcup N_3}Q_i+\frac{3ac}{4}+\frac{ac}{4}\\&> a\sum _{i\in N_2\bigcup N_3}Q_i+\frac{3a}{4}\left( P_{1l}^{\sigma _1}+\sum _{i\in N_3}Q_i \right) +\frac{a}{2}\sum _{i\in N_3}Q_i\\&\ge \frac{3a}{2} \left[ \frac{1}{2} \left( P_{1l}^{\sigma _1}+\sum _{i\in N_2}Q_i \right) +\frac{3}{2}\sum _{i\in N_3}Q_i \right] \end{aligned}$$

and $\frac{f(\pi _{14})}{f^*}\le \frac{5}{3}$.

Case 5.3.2: If $S_{\Delta _2}\ne \emptyset$ and $\sigma _1[2]$ is completed on day 3 in schedule $\pi _{14}$, then $\sum _{i\in N_3}<P_{2r}^{\sigma _1}$. By the same analysis as that for case 5.2, there must be one order $i$ with $Q_i\ge Q_{\sigma _1[2]}$ being completed on day 3. Since there must be one order $i\in \{\sigma _1[1]\}\bigcup \Delta _1$ with $Q_i\ge Q_{\sigma _1[1]}$ being completed on day 2,

$$\begin{aligned} f^*>a\sum _{i\in N_2\bigcup N_3}Q_i+aQ_{\sigma _1[1]}+2aQ_{\sigma _1[2]}. \end{aligned}$$

By $\sum _{i\in N_3}<P_{2r}^{\sigma _1}$ and $\sum _{i\in N_2\bigcup N_3}<P_{1r}^{\sigma _1}$, we have

$$\begin{aligned} f^*>2a\sum _{i\in N_2}Q_i+3a\sum _{i\in N_3}Q_i+aP_{1l}^{\sigma _1}+2aP_{2l}^{\sigma _1}. \end{aligned}$$

By the same analysis as that of case 5.3.1,

$$\begin{aligned} f(\pi _{14})-f(\sigma _0)<\frac{a}{2} \left( P_{1l}^{\sigma _1}+\sum _{i\in N_2}Q_i \right) +\frac{3a}{2}\sum _{i\in N_3}Q_i+aP_{2l}^{\sigma _1} \end{aligned}$$

and $\frac{f(\pi _{14})}{f^*}\le \frac{5}{3}$.

Case 5.3.3: If $S_{\Delta _2}=\emptyset$, we consider the following four subcases:

Case 5.3.3.1: If $\Delta _2\ne \emptyset$ and $\sigma _1[2]$ is completed on day 2 in schedule $\pi _{14}$, note that $\sum _{i\in N_2\bigcup N_3}Q_i<P_{1r}^{\sigma _1}$, then $Q_{\Delta _2,max}>P_{1l}^{\sigma _1}+\sum _{i\in N_2\bigcup N_3}Q_i$ and

$$\begin{aligned} f^*>a\sum _{i\in N_2\bigcup N_3}Q_i+aQ_{\sigma _1[1]}+aQ_{\Delta _2,max}>3a\sum _{i\in N_2\bigcup N_3}Q_i+2aP_{1l}^{\sigma _1}. \end{aligned}$$

By then same analysis as that of case 5.3.1, we have

$$\begin{aligned} f(\pi _{14})-f(\sigma _0)<a \left( P_{1l}^{\sigma _1}+\sum _{i\in N_2}Q_i \right) +2a\sum _{i\in N_3}Q_i \end{aligned}$$

and $\frac{f(\pi _{14})}{f^*}\le \frac{5}{3}$.

Case 5.3.3.2: If $\Delta _2\ne \emptyset$ and $\sigma _1[2]$ is completed on day 3 in schedule $\pi _{14}$, note that either $O_{\Delta _2,max}$ or $\sigma _1[1]$ can not be completed on day 1 and there must be one order $i$ with $Q_i\ge Q_{\sigma _1[2]}$ being completed on day 3, therefore,

$$\begin{aligned} f^*&> a\sum _{i\in N_2\bigcup N_3}Q_i+aQ_{\sigma _1[1]}+aQ_{\Delta _2,max}\\&+\,2aQ_{\sigma _1[2]}>3a\sum _{i\in N_2\bigcup N_3}Q_i+2aP_{1l}^{\sigma _1}+2aP_{2l}^{\sigma _1}. \end{aligned}$$

By the same analysis as that of case 5.3.1, we have

$$\begin{aligned} f(\pi _{14})-f(\sigma _0)<a \left( P_{1l}^{\sigma _1}+\sum _{i\in N_2}Q_i \right) +2a\sum _{i\in N_3}Q_i+aP_{2l}^{\sigma _1} \end{aligned}$$

and $\frac{f(\pi _{14})}{f^*}\le \frac{5}{3}$.

Case 5.3.3.3: If $\Delta _2=\emptyset$ and $\sigma _1[2]$ is completed on day 2 in schedule $\pi _{14}$, it is then clear to see that $\sigma _1[1]$ is completed on day 2 in schedule $\pi _{15}$ and $f(\pi _{15})-f^*<a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i$. By condition (2) and $\sum _{i\in N_2\bigcup N_3}<P_{1r}^{\sigma _1}$, we have

$$\begin{aligned} 2c-\sum _{i\in N_2\bigcup N_3}Q_i-P_{2r}^{\sigma _1}\ge P_{1l}^{\sigma _{1}}+P_{2l}^{\sigma _{1}}>\frac{2}{3}\left( \sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+c \right) \end{aligned}$$

and $c>2\sum _{i\in N_2\bigcup N_3}Q_i$. Thus,

$$\begin{aligned} f^*>ac+a\sum _{i\in N_2\bigcup N_3}Q_i>3a\sum _{i\in N_2\bigcup N_3}Q_i \end{aligned}$$

and $\frac{f(\pi _{15})}{f^*}\le \frac{5}{3}$.

Case 5.3.3.4: If $\Delta _2=\emptyset$ and $\sigma _1[2]$ is completed on day 3 in schedule $\pi _{14}$, by the same analysis as that of case 5.3.2, we have

$$\begin{aligned} f^*&> a\sum _{i\in N_2\bigcup N_3}Q_i+aQ_{\sigma _1[1]}+2aQ_{\sigma _1[2]}>2a\sum _{i\in N_2}Q_i\\&+\,3a\sum _{i\in N_3}Q_i+aP_{1l}^{\sigma _1}+2aP_{2l}^{\sigma _1} \end{aligned}$$

and

$$\begin{aligned} f(\pi _{14})-f(\sigma _0)<\frac{a}{2} \left( P_{1l}^{\sigma _1}+\sum _{i\in N_2}Q_i \right) +\frac{3a}{2}\sum _{i\in N_3}Q_i+aP_{2l}^{\sigma _1}. \end{aligned}$$

Therefore, $\frac{f(\pi _{14})}{f^*}\le \frac{5}{3}$. $\square$

Theorem 3.3 If the output schedule of heuristic $\mathbf {H}$ is obtained from step 3, then $\frac{f(\sigma ^{\mathbf {H}})}{f^*}\le 5/3$.

Proof

This is a direct conclusion from Lemmas 6.1 through 6.5. $\square$

Appendix C: Proof of Theorem 3.4

Lemma 7.1

If the heuristic runs step 4 and the obtained schedule is $\sigma _1$, then $\frac{f(\sigma ^{\mathbf {H}})}{f^*}\le \frac{5}{3}$.

Proof

In schedule $\sigma _1$, it is clear to see that

$$\begin{aligned} f(\sigma _1)-f(\sigma _0)=a\sum _{i=1}^{3}P_{il}^{\sigma _1}~~~~and~~~~f^*\ge f(\sigma _0). \end{aligned}$$

(1) If $d_{\sigma _{1}[2]}\le d_{max}-1$, by $b-a(d_{max}-1)>0$ and EDD-LPT, we have

$$\begin{aligned} f(\sigma _0)&> c[b-a(d_{\sigma _1[2]}-1)]+c[b-a(d_{\sigma _1[2]}-2)]+c[b-a(d_{max}-3)]\\&> ac+2ac+2ac\\&= 5ac \end{aligned}$$

and $\frac{f(\sigma _1)}{f^*}\le 1+\frac{f(\sigma _1)-f(\sigma _0)}{f(\sigma _0)}=1+\frac{\sum _{i=1}^{3}P_{il}^{\sigma _1}}{f(\sigma _0)}<1+\frac{3ac}{5ac}<\frac{5}{3}.$

(2) If $d_{\sigma _{1}[1]}\le d_{max}-2$, then we have

$$\begin{aligned} f(\sigma _0)&> c[b-a(d_{\sigma _1[1]}-1)]+c[b-a(d_{max}-2)]+c[b-a(d_{max}-3)]\\&> 2ac+ac+2ac\\&= 5ac \end{aligned}$$

and $\frac{f(\sigma _1)}{f^*}\le 1+\frac{f(\sigma _1)-f(\sigma _0)}{f(\sigma _0)}=1+\frac{\sum _{i=1}^{3}P_{il}^{\sigma _1}}{f(\sigma _0)}<1+\frac{3ac}{5ac}<\frac{5}{3}.$

(3) If $d_{max}=d_{\sigma _{1}[1]}+1$ and $\sum _{i=1}^{3}P_{il}^{\sigma _{1}}\le \frac{2}{3}(4c+3\Delta x)$, then

$$\begin{aligned} f(\sigma _0)&> c[b-a(d_{\sigma _1[1]}-1)]+c[b-a(d_{max}-2)]+c[b-a(d_{max}-3)]\\&+\,\Delta x[b-a(d_{max}-4)]\\&> ac+ac+2ac+3\Delta x\\&= a(4c+3\Delta x) \end{aligned}$$

and

$$\begin{aligned} \frac{f(\sigma _1)}{f^*}&\le 1+\frac{f(\sigma _1)-f(\sigma _0)}{f(\sigma _0)}\\&= 1+\frac{a\sum _{i=1}^{3}P_{il}^{\sigma _1}}{f(\sigma _0)}\\&< 1+\frac{\frac{2}{3}a(4c+3\Delta x)}{a(4c+3\Delta x)}\\&= \frac{5}{3}. \end{aligned}$$

(4) If $d_{max}=d_{\sigma _{1}[1]}$ and $\sum _{i=1}^{3}P_{il}^{\sigma _{1}}\le \frac{2}{3}\big (3\sum _{i\in N_1}Q_i+2\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+\sum _{i\in N_4}Q_i+3c+3\Delta x \big )$, we have

$$\begin{aligned} f(\sigma _0)&> \sum _{i\in N_1}Q_i[b-a(2-1)]+\sum _{i\in N_2}Q_i[b-a(3-1)]+\sum _{i\in N_3}Q_i[b-a(4-1)]\\&+\,\sum _{i\in N_4}Q_i[b-a(5-1)]+c[b-a(d_{max}-2)]+c[b-a(d_{max}-3)]\\&+\,\Delta x[b-a(d_{max}-4)]\\&> a\left( 3\sum _{i\in N_1}Q_i+2\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+\sum _{i\in N_4}Q_i+3c+3\Delta x \right) \end{aligned}$$

and

$$\begin{aligned} \frac{f(\sigma _1)}{f^*}&\le 1+\frac{f(\sigma _1)-f(\sigma _0)}{f(\sigma _0)}\\&= 1+\frac{a\sum _{i=1}^{3}P_{il}^{\sigma _1}}{f(\sigma _0)}\\&< 1{+}\frac{\frac{2}{3}a(3\sum _{i\in N_1}Q_i{+}2\sum _{i\in N_2}Q_i{+}\sum _{i\in N_3}Q_i{+}\sum _{i\in N_4}Q_i{+}3c}{a(3\sum _{i\in N_1}Q_i{+}2\sum _{i\in N_2}Q_i{+}\sum _{i\in N_3}Q_i{+}\sum _{i\in N_4}Q_i{+}3c}\frac{{+}3\Delta x)}{{+}3\Delta x )}\\&= \frac{5}{3}. \end{aligned}$$

We can conclude that, if the output schedule is $\sigma _1$, then $\frac{f(\sigma _1)}{f^*}\le \frac{5}{3}$. $\square$

Lemma 7.2

If the heuristic runs step 4 and the output schedule is $\sigma _2^{P4}$, then $\frac{f(\sigma _2^{P4})}{f^*}{\le } \frac{5}{3}$.

Proof

Let $N_{5}=\{i\in N|d_{i}=d_{\sigma _1[1]}\}\backslash \{\sigma _1[1]\}$, if $d_{max}\ge d_{\sigma _1[1]}+1$. In this case, the following three conditions hold:

• (1) $d_{\sigma _1[1]}=d_{max}-1=4$;

• (2) $\sum _{i=1}^{3}P_{il}^{\sigma _{1}}>\frac{2}{3}(4c+3\Delta x)$;

• (3) $d_{\sigma _1[2]}=d_{max}=5$.

By condition (2) and $Q_i<c$, we have $P_{il}^{\sigma _{1}}>\frac{2}{3}c+2\Delta x$, for $i=1 , 2, 3$, and $Q_{\sigma _1[i]}>\frac{2}{3}c+2\Delta x$, for $i=1 , 2, 3$. The structure of $\sigma _1$(and $\sigma _0$) is $(N_1, N_2, \sigma _1[1], N_5, \sigma _1[2], \sigma _1[3], N_6).$ By the definitions of $N_1$ and $N_2$, the orders in $N_1$ must be finished production by the end of day 1 and the orders in $N_2$ must be finished production by the end of day 2. By the definition of $N_5$ and condition (1), the orders in $N_5\bigcup \{\sigma _1[1]\}$ must be finished production by the end of day 3. By $Q_{\sigma _1[i]}>\frac{2}{3}c+2\Delta x$, for $i=1 , 2, 3$, we can see that at most one order in $\{\sigma _1[1], \sigma _1[2], \sigma _1[3]\}$ is finished on day 1 in any feasible schedule. Since $\sum _{i\in N_1\bigcup N_2\bigcup N_5}Q_i+\sum _{i=1}^{3}Q_{\sigma _1[i]}>3c$, at least one order in $\{\sigma _1[2], \sigma _1[3]\}$ is finished on day 4 in any feasible schedule. Therefore,

$$\begin{aligned} f(\sigma _2^{P4})-f^*\le a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_5}Q_i+3a\sum _{i\in N_6}Q_i. \end{aligned}$$

Note that $\sum _{i\in N_2\bigcup N_5\bigcup N_6}Q_i<\frac{1}{3}c$ and $f(\sigma _0)>a(4c+3\Delta x)$, then

$$\begin{aligned} \frac{f(\sigma _2^{P4})}{f^*}&\le 1+\frac{f(\sigma _2^{P4})-f^*}{f(\sigma _0)}\\&< 1+\frac{a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_5}Q_i+3a\sum _{i\in N_6}}{a(4c+3\Delta x)}\\&< 1+\frac{ac}{4ac}\\&< \frac{5}{3}. \end{aligned}$$

$\square$

Lemma 7.3

If the heuristic runs step 4 and the output schedule is $\sigma _3^{P4}$, then $\frac{f(\sigma _3^{P4})}{f^*}\le \frac{5}{3}$.

Proof

We define:

• $\Delta _1$ $=$  the set of orders between $N_4$ and $\sigma _1[1]$ in schedule $\sigma _1$.

• $\Delta _2$ $=$ the set of orders between $\sigma _1[1]$ and $\sigma _1[2]$ in schedule $\sigma _1$.

• $N_{6}$ $=$ the set of the orders processed after order $\sigma _{1}[3]$ in $\sigma _1$.

• $S=N_2\bigcup N_3\bigcup N_4\bigcup N_6$.

We first consider case (1): $d_{max}=d_{\sigma _{1}[1]}+1$, $\sum _{i=1}^{3}P_{il}^{\sigma _{1}}>\frac{2}{3}(4c+3\Delta x)$ and $d_{max}\ge 6$. As discussed in the proof of Lemma 7.2, the structure of $\sigma _1$ (and $\sigma _0$) is $(N_1, N_2, N_3, \sigma _1[1], N_5,$ $\sigma _1[2], \sigma _1[3], N_6)$ and $Q_{\sigma _1[i]}>\frac{2}{3}c+2\Delta x$, for $i=1 , 2, 3$. By the same analysis as that in the proof of Lemma 7.2, we can see that at most one order from $\{\sigma _1[1], \sigma _1[2], \sigma _1[3]\}$ is started and completed in the same day in any feasible schedule, then

$$\begin{aligned} f(\sigma _3^{P4})-f^*\le a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+3a\sum _{i\in N_5\bigcup N_6}Q_i. \end{aligned}$$

Since $\sum _{i=1}^{3}P_{il}^{\sigma _{1}}>\frac{2}{3}(4c+3\Delta x)$, $\sum _{i\in S}Q_i<\frac{1}{3}c$. Note that $f(\sigma _0)>a(4c+3\Delta x)$, then

$$\begin{aligned} \frac{f(\sigma _3^{P4})}{f^*}&\le 1+\frac{f(\sigma _3^{P4})-f^*}{f(\sigma _0)}\\&< 1+\frac{a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+3a\sum _{i\in N_5\bigcup N_6}Q_i}{a(4c+3\Delta x)}\\&< 1+\frac{ac}{4ac}\\&< \frac{5}{3}. \end{aligned}$$

Next we consider case (2): $d_{max}=d_{\sigma _{1}[1]}$, $\sum _{i=1}^{3}P_{il}^{\sigma _{1}}>\frac{2}{3} \big (3\sum _{i\in N_1}Q_i+2\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+\sum _{i\in N_4}Q_i+3c+3\Delta x \big )$ and $P_{3l}^{\sigma _1}\ge \frac{c}{2}$. Schedule $\sigma _{1}$ has the structure of $(N_1, N_2, N_3,$ $N_4, \Delta _1, \sigma _{1}[1], \Delta _2,$ $\sigma _{1}[2], \Delta _3, \sigma _{1}[3], N_6)$ and the following three conditions hold:

• (1) $d_{max}=d_{\sigma _{1}[1]}$;

• (2) $\sum _{i=1}^{3}P_{il}^{\sigma _{1}}>\frac{2}{3}\big (3\sum _{i\in N_1}Q_i+2\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+\sum _{i\in N_4}Q_i+3c+3\Delta x \big )$;

• (3) $P_{3l}^{\sigma _1}\ge \frac{c}{2}$.

Based on these conditions, we have the following conclusions:

Result 1: $Q_{\sigma _1[1]}>\frac{2}{3}c$. By condition $\sum _{i=1}^{3}P_{il}^{\sigma _{1}}>2c+2\Delta x$ and $Q_i<c$, for $i\in N$, we have $P_{max,l}^{\sigma _1}=$max$\{P_{il}^{\sigma _1}|$ $i=1, 2, 3\}>\frac{2}{3}(c+\Delta x)$. By EDD-LPT, we have $Q_{\sigma _1[1]}>\frac{2}{3}c$.

Result 2: $\Delta _1=\emptyset$. If $\Delta _1\ne \emptyset$, we have $d_j=d_{max}$ and $Q_j\ge Q_{\sigma _1[1]}>\frac{2}{3}c$ for $j\in \Delta _1$ (By EDD-LPT). Thus, $P_{1l}^{\sigma _1}<\frac{1}{3}c$. Because $Q_{\sigma _1[1]}>\frac{2}{3}c$, $P_{1r}^{\sigma _1}>\frac{1}{3}c$. By $\sum _{i=1}^{3}P_{il}^{\sigma _{1}}>2c+2\Delta x$, we have $P_{2l}^{\sigma _1}+P_{3l}^{\sigma _1}>\frac{5}{3}c$ and $P_{2l}^{\sigma _1}>\frac{2}{3}c$. Thus $P_{1r}^{\sigma _1}+P_{2l}^{\sigma _1}>c$, which contradicts with the available daily production capacity of day 2. Therefore, $\Delta _1=\emptyset$.

Result 3: $\Delta _3=\emptyset$. If $\Delta _3\ne \emptyset$, $d_j=d_{max}$ and $Q_j\ge Q_{\sigma _1[3]}>\frac{1}{2}c$ for $j\in \Delta _3$ (By EDD-LPT). By $P_{3l}^{\sigma _1}\ge \frac{c}{2}$, we have $Q_j+P_{3l}^{\sigma _1}>c$ for $j\in \Delta _3$, which contradicts with the available daily production capacity of day 3. Therefore, $\Delta _3=\emptyset$.

Result 4: $\Delta _2=\emptyset$. If $\Delta _2\ne \emptyset$, we have $d_j=d_{max}$, $Q_j\ge Q_{\sigma _1[3]}>\frac{1}{2}c$ and $P_{2l}^{\sigma _1}<\frac{c}{2}$ for $j\in \Delta _2$ (By EDD-LPT). Consider the following two cases:

Case (i) If $P_{3l}^{\sigma _1}\ge \frac{2c}{3}$, then $Q_j\ge Q_{\sigma _1[2]}\ge Q_{\sigma _1[3]}>\frac{2c}{3}$ for $j\in \Delta _2$. By the same analysis as that for Result 2, $\Delta _2=\emptyset$.

Case (ii) If $\frac{c}{2}\le P_{3l}^{\sigma _1}<\frac{2c}{3}$, then $P_{1l}^{\sigma _1}+P_{2l}^{\sigma _1}>\frac{4c}{3}$ and $P_{2l}^{\sigma _1}>\frac{c}{3}$. By Result 3, we have $P_{2r}^{\sigma _1}>\frac{c}{3}$ and $Q_{\sigma _1[2]}=P_{2l}^{\sigma _1}+P_{2r}^{\sigma _1}>\frac{2c}{3}$. Thus, $Q_j\ge Q_{\sigma _1[2]}>\frac{2c}{3}$ for $j\in \Delta _2$, and by the same analysis as that for Result 2, $\Delta _2=\emptyset$.

Based on the above analysis, we can see that the structure of $\sigma _1$ (and $\sigma _0$) is $(N_1, N_2, N_3, N_4, \sigma _{1}[1],$ $\sigma _{1}[2], \sigma _{1}[3], N_6)$. Since $\sum _{i=1}^{3}P_{il}^{\sigma _{1}}>\frac{2}{3}(3\sum _{i\in N_1}Q_i+2\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+\sum _{i\in N_4}Q_i+3c+3\Delta x)$,

$$\begin{aligned} \sum _{i\in S}Q_i&< \sum _{i\in N}Q_i-\sum _{i=1}^{3}P_{il}^{\sigma _1}\\&< c-\Delta x-\frac{2}{3}\left( \sum _{i\in N_1}Q_i+\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+\sum _{i\in N_4}Q_i \right) \\&< c-\frac{2}{3}\left( \sum _{i\in S}Q_i \right) \end{aligned}$$

and $\sum _{i\in S}Q_i<\frac{3c}{5}$.

As discussed in the proof of Lemma 7.2, we have

$$\begin{aligned} f(\sigma _0)>a \left( 3\sum _{i\in N_1}Q_i+2\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+\sum _{i\in N_4}Q_i+3c+3\Delta x \right) >3ac \end{aligned}$$

and

$$\begin{aligned} f(\sigma _3^{P4})-f^*\le a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+3a\sum _{i\in N_4\bigcup N_6}Q_i. \end{aligned}$$

Thus,

$$\begin{aligned} \frac{f(\sigma _3^{P4})}{f^*}&\le 1+\frac{f(\sigma _3^{P4})-f^*}{f(\sigma _0)}\\&< 1+\frac{a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+3a\sum _{i\in N_4\bigcup N_6}Q_i}{3ac}\\&< 1+\frac{\frac{9c}{5}}{3ac}\\&< \frac{5}{3}. \end{aligned}$$

$\square$

Lemma 7.4

If the heuristic runs step 4 and the output schedule is $\sigma _4^{P4}$, then $\frac{f(\sigma _4^{P4})}{f^*}\le \frac{5}{3}$.

Proof

In this case, the following three conditions hold:

• (1) $d_{max}=d_{\sigma _{1}[1]}$;

• (2) $\sum _{i=1}^{3}P_{il}^{\sigma _{1}}>\frac{2}{3}\big (3\sum _{i\in N_1}Q_i+2\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+\sum _{i\in N_4}Q_i+3c+3\Delta x \big )$;

• (3) $P_{3l}^{\sigma _1}<\frac{c}{2}$.

Based on these conditions, we have the following conclusions:

Result 1: $\Delta _1=\emptyset$. As discussed in the proof of Lemma 7.3, we have $\Delta _1=\emptyset$.

Result 2: $\Delta _2=\emptyset$. If $\Delta _2\ne \emptyset$, then $d_j=d_{max}$, $Q_j\ge Q_{\sigma _1[2]}$ and $P_{2l}^{\sigma _1}<\frac{c}{2}$ for $j\in \Delta _2$ (By EDD-LPT). By condition (2), we can see that $P_{1l}^{\sigma _1}+P_{3l}^{\sigma _1}>\frac{3}{2}c$. By condition (3), we have $P_{1l}^{\sigma _1}>c$, which contradicts with $Q_i<c$ for $i\in N$. Thus, $\Delta _2=\emptyset$.

Result 3: $\Delta _3\ne \emptyset$. If $\Delta _3=\emptyset$, by condition (3) and $P_{2r}^{\sigma _1}+P_{3l}^{\sigma _1}=c$, we have $P_{2r}^{\sigma _1}>\frac{c}{2}$ and $P_{2l}^{\sigma _1}<\frac{c}{2}$. By condition (2), we have $P_{1l}^{\sigma _1}+P_{3l}^{\sigma _1}>\frac{3}{2}c$, and by condition (3), we have $P_{1l}^{\sigma _1}>c$, which contradicts with $Q_i<c$ for $i\in N$. Therefore, $\Delta _3\ne \emptyset$.

Result 4: $P_{3l}^{\sigma _1}>\frac{5}{3}\sum _{i\in E}Q_i+2\Delta x$ (Let $E=N_1\bigcup N_2\bigcup N_3\bigcup N_4$). By condition (2) and $P_{1l}^{\sigma _1}+\sum _{i\in E}Q_i=c$, we have

$$\begin{aligned} c-\sum _{i\in E}Q_i+c+P_{3l}^{\sigma _1}&> P_{1l}^{\sigma _1}+P_{2l}^{\sigma _1}+P_{3l}^{\sigma _1}\\&> \frac{2}{3}\left( 3\sum _{i\in N_1}Q_i{+}2\sum _{i\in N_2}Q_i{+}\sum _{i\in N_3}Q_i{+}\sum _{i\in N_4}Q_i{+}3c{+}3\Delta x \right) \\&> \frac{2}{3}\left( \sum _{i\in E}Q_i+3c+3\Delta x \right) \end{aligned}$$

and $P_{3l}^{\sigma _1}>\frac{5}{3}\sum _{i\in E}Q_i+2\Delta x$.

Result 5: $\sum _{i\in \Delta _3}Q_i<P_{1l}^{\sigma _1}<P_{2l}^{\sigma _1}$. It is clear to see that

$$\begin{aligned} \sum _{i\in \Delta _3}Q_i+P_{3l}^{\sigma _1}<c=P_{1l}^{\sigma _1}+\sum _{i\in E}Q_i=P_{1r}^{\sigma _1}+P_{2l}^{\sigma _1}. \end{aligned}$$

By $P_{1l}^{\sigma _1}+P_{1r}^{\sigma _1}=Q_{\sigma _1[1]}<c$ and Result 4, we have $\sum _{i\in \Delta _3}Q_i<P_{1l}^{\sigma _1}<P_{2l}^{\sigma _1}$.

Result 6: $\sigma _{1}[3]$ is the only order that is finished production on day 4 in $\sigma _4^{P4}$. This is implied by Result 4.

From the above analysis, we can see that the structure of $\sigma _1$ (and $\sigma _0$) is $(N_1, N_2, N_3, N_4, \sigma _{1}[1],$ $\sigma _{1}[2], \Delta _3, \sigma _{1}[3], N_6)$. By the definition of $N_i$, the orders in $N_i$ must be finished production by the end of day $i$, for $i=1,2,3$.

Now we consider three possible cases:

Case (4.1) If at least one order in $\Delta _3$ is finished production on day 1 in the optimal schedule, by the fact that $d_j=d_{max}$, $Q_j\ge Q_{\sigma _3[1]}$ for $j\in \Delta _3$ (By EDD-LPT) and Result 4, we have $Q_j+Q_{\sigma _1[1]}+Q_{\sigma _1[2]}>2c$ and $Q_j+$min$\{Q_{\sigma _1[1]},Q_{\sigma _1[2]}\}>c$ for $j\in \Delta _3$. Furthermore, none of $\{Q_{\sigma _1[1]},Q_{\sigma _1[2]}\}$ can be finished on day 1, at most one order of $\{Q_{\sigma _1[1]},Q_{\sigma _1[2]}\}$ is completed by day 2 and at least one order in $\{Q_{\sigma _1[1]},Q_{\sigma _1[2]}\}$ is completed by day 3 in the optimal schedule. By Result 5, we have

$$\begin{aligned} \sum _{i\in N_1\bigcup \Delta _3\bigcup N_2}Q_i<c,~~\sum _{i\in N_1\bigcup \Delta _3\bigcup N_2}Q_i+Q_{\sigma _1[1]}+Q_{\sigma _1[2]}<3c \end{aligned}$$

and

$$\begin{aligned} \sum _{i\in N_1\bigcup \Delta _3\bigcup N_2}Q_i+max\{Q_{\sigma _1[1]},Q_{\sigma _1[2]}\}<2c. \end{aligned}$$

If $\sigma _1[1]$ is processed before $\sigma _1[2]$ in the optimal schedule, we have

$$\begin{aligned} f(\pi _{16})-f^*<a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+2a\sum _{i\in N_4\bigcup N_6}Q_i+3aQ_{\sigma _1[3]}, \end{aligned}$$

and if $\sigma _1[2]$ is processed before $\sigma _1[1]$ in the optimal schedule, we have

$$\begin{aligned} f(\pi _{15})-f^*<a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+2a\sum _{i\in N_4\bigcup N_6}Q_i+3aQ_{\sigma _1[3]}. \end{aligned}$$

Therefore,

$$\begin{aligned} f(\sigma _4^{P4})-f^*<a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+2a\sum _{i\in N_4\bigcup N_6}Q_i+3aQ_{\sigma _1[3]}. \end{aligned}$$

Case (4.2) If no order in $\Delta _3$ is finished production on day 1 and at least one order in $\Delta _3$ is finished production on day 2 in the optimal schedule, by the same analysis as that for case (4.1), we can see that at least one order of $\{Q_{\sigma _1[1]},Q_{\sigma _1[2]}\}$ is finished production on day 3 in the optimal schedule. By the same analysis as that for case (4.1), if $\sigma _1[1]$ is processed before $\sigma _1[2]$ in the optimal schedule, then

$$\begin{aligned} f(\pi _{12})-f^*<a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+2a\sum _{i\in N_4\bigcup N_6}Q_i+3aQ_{\sigma _1[3]}, \end{aligned}$$

and if $\sigma _1[2]$ is processed before $\sigma _1[1]$ in the optimal schedule, then

$$\begin{aligned} f(\pi _{14})-f^*<a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+2a\sum _{i\in N_4\bigcup N_6}Q_i+3aQ_{\sigma _1[3]}. \end{aligned}$$

Therefore,

$$\begin{aligned} f(\sigma _4^{P4})-f^*<a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+2a\sum _{i\in N_4\bigcup N_6}Q_i+3aQ_{\sigma _1[3]}. \end{aligned}$$

Case (4.3) If the orders in $\Delta _3$ are finished production after day 2, by the same analysis as that for case (4.1), if $\sigma _1[1]$ is processed before $\sigma _1[2]$ in the optimal schedule,

$$\begin{aligned} f(\pi _{11})-f^*<\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+2a\sum _{i\in N_4\bigcup N_6}Q_i+3aQ_{\sigma _1[3]}, \end{aligned}$$

and if $\sigma _1[2]$ is processed before $\sigma _1[1]$ in the optimal schedule,

$$\begin{aligned} f(\pi _{13})-f^*<a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+2a\sum _{i\in N_4\bigcup N_6}Q_i+3aQ_{\sigma _1[3]}. \end{aligned}$$

Therefore,

$$\begin{aligned} f(\sigma _4^{P4})-f^*<a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i+2a\sum _{i\in N_4\bigcup N_6}Q_i+3aQ_{\sigma _1[3]}. \end{aligned}$$

Note that

$$\begin{aligned} f(\sigma _0)>a\left( 3\sum _{i\in N_1}Q_i+2\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+\sum _{i\in N_4}Q_i+3c+3\Delta x\right) , \end{aligned}$$

then

$$\begin{aligned}&2f(\sigma _0)-3(f(\sigma _4^{P4})-f^*)\\&\quad >2a \left( 3\sum _{i\in N_1}Q_i+2\sum _{i\in N_2}Q_i+\sum _{i\in N_3}Q_i+\sum _{i\in N_4}Q_i+3c+3\Delta x\right) \\&\qquad -\,3\left[ a\sum _{i\in N_2}Q_i+2a\sum _{i\in N_3}Q_i\right. \\&\qquad +\left. 2a\sum _{i\in N_4\bigcup N_6}Q_i+3aQ_{\sigma _1[3]}\right] \\&\quad =6ac+6a\Delta x+aN_2-4a\sum _{i\in N_3\bigcup N_4}Q_i-6a\sum _{i\in N_6}Q_i-9aQ_{\sigma _1[3]}\\&\quad >6ac+9aP_{3r}^{\sigma _1}-9aQ_{\sigma _1[3]}-4a\sum _{i\in N_3\bigcup N_4}Q_i-3a\Delta x\\&\quad >6ac-9aP_{3l}^{\sigma _1}-3a\Delta x-4a\sum _{i\in N_3\bigcup N_4}Q_i\\&\quad >\frac{3}{2}ac-4a\sum _{i\in N_3\bigcup N_4}Q_i-3a\Delta x. \end{aligned}$$

By Result 4 and condition (3), we have $\sum _{i\in E}Q_i+\Delta x<\frac{3c}{10}$ and

$$\begin{aligned} 2f^*-3(f(\sigma _4^{P4})-f^*)>\frac{3}{2}ac-4a\sum _{i\in N_3\bigcup N_4}Q_i-3a\Delta x>\frac{3}{2}ac-\frac{6}{5}ac>0. \end{aligned}$$

Therefore, if the output schedule is $\sigma _4^{P4}$, then $\frac{f(\sigma _4^{P4})}{f^*}\le \frac{5}{3}$. $\square$

Theorem 3.4. If the output schedule of heuristic $\mathbf {H}$ is obtained from step 4, then $\frac{f(\sigma ^{\mathbf {H}})}{f^*}\le 5/3$.

Proof

This is a direct conclusion from Lemmas 7.1 through 7.4. $\square$